You visited us 1 times! Enjoying our articles? Unlock Full Access!

Centripetal Force

A thin unifor...

Question

A thin uniform rod $A B$ of mass $m = 1.0 k g$ moves translationally with acceleration $w = 2.0 m / s^{2}$ due to two antiparallel forces $F_{1}$ and $F_{2}$ (figure shown above). The distance between the points at which these forces are applied is equal to $a = 20 c m$ . Besides, it is known that $F_{2} = 5.0 N$ . Find the length of the rod in metres.

Open in App

Solution

Since, motion of the rod is purely translational, net torque about the C.M. of the rod should be equal to zero.
Thus, $F_{1} \frac{l}{2} = F_{2} (\frac{l}{2} - a)$ or,
$\frac{F_{1}}{F_{2}} = 1 - \frac{a}{\frac{l}{2}}$ ....(1)
For the translational motion of rod,
$F_{2} - F_{1} = m w_{c}$ or, $1 - \frac{F_{1}}{F_{2}} = \frac{m w_{c}}{F_{2}}$ ..... (2)
From (1) and (2)
$\frac{a}{\frac{l}{2}} = \frac{m w_{c}}{F_{2}}$
or, $l = \frac{2 a F_{2}}{m w_{c}} = 1 m$

Suggest Corrections

Similar questions

1 k g

a = 2 m / s^{2}

F_{1}

F_{2}

ℓ = 0.2 m

F_{1} = 8 N

A B

m

a

F_{1}

F_{2}

b

I

F

L

M

y

F_{1}

F_{1}

F_{2}

F_{2} < F_{1}

L

F_{1}

F_{2}

F_{2} > F_{1}

F

l

F_{1}

Join BYJU'S Learning Program